Monday, 18 March 2019

Well, this was a pain in the backside to edit. The film is so tawdry and dull that we kept getting lost on tangents. Fear not though faithful listener, Thomas has edited the two hours of guff down to a single hour of solid... bronze.

Friday, 15 March 2019

one person randomly decides whether
to tell the truth or lie (assume lies and truth are equally likely);

the three know amongst themselves who they are.

You can ask two questions to the people. The answer to which
must either be yes or no. What question do you ask and who do you ask?

This is an extension of the famous two person puzzle. Normally, you only have two guards, one tells the truth and one lies. You have to choose and open one of the doors, but you can only ask a single question to one of the guards.

What do you ask so you can pick the door to freedom?

In this case the solution is:If I asked what door would lead to freedom, what door would the other guard point to?

This works by considering the two possible outcomes. Namely:

If you asked the truth-guard, the truth-guard would tell you that the liar-guard would point to the door that leads to death.

If you asked the liar-guard, the liar-guard would tell you that the truth-guard would point to the door that leads to death.

Therefore, no matter who you ask, the guards tell you which door leads to death, and therefore you can pick the other door.

The inclusion of the trickster guard, however, changes the puzzle dramatically. Specifically, you questions have to work no matter who is being asked (truth-teller, liar, or trickster). Further, no matter what you ask, you always
have to worry about the trickster screwing up your logic.

Thus, one strategy is to identify one person is NOT the trickster. We
don't have to identify whether they are truth-teller, or liar.

Call the three gaurds A, B and C. You ask A:

"Is exactly one of these statements true:

You are the truth-teller

B is the trickster

If you get back the answer yes, then the possibilities are:

A is the truth-teller and B is the liar (1. true, 2. false, so one statement true, so answer is yes which truth-teller truthfully gives)

A is the trickster

A is the liar and B is the truth-teller (both statements false so answer is no which liar lies about)

In all three cases, B is not the trickster.

If you get back the answer no, then the possibilities are:

A is the truth-teller and B is the trickster (both statements true, so answer is no which truth-teller truthfully gives)

A is the trickster

A is the liar and B is the trickster (1. false, 2. true so one statement true so answer is yes which liar lies about)

In all three cases, C is not the trickster.

Once you have found a person who is not the trickster, just point to a door and ask the person:

Tuesday, 19 February 2019

Welcome to the strangely erotic episode of Maths at, where we watch the tense, psychological thriller, Fermat's Room (or La Habitación de Fermat, for you Spanish speakers) and we ask the real questions of... WHAT HAPPENED ON THE BOAT?

As per usual, the time line is all wonky. This episode does follow on from A Beautiful Mind, but was recorded a long time after, so although we talk about our lives having changed dramatically, it's only bee two weeks for you and you already know what's happened if you've listened to our Christmas episode. It's so hard living in a linear timeline.

So if you want to know:

what Liz's ovaries sound like;

which superpower our hosts would rather have;

how Ben would overhaul examination procedures,

then join us in our latest episode of being distracted by pop corn makers.

If you're interested in watching Fermat's Room and want an easier time than we had in finding it, simply click the Amazon link below.

Friday, 8 February 2019

Two trains are on the same track. They start 100km apart and head towards each other at a speed of 50km/h.

Whilst
these two trains are heading for their collision a fly starts out on
the front of one train and zooms directly to the front of the other at a
speed of 75km/h (see the animation above). Once the fly reaches the
second train it immediately darts back to the front of the first train
at the same speed and repeats this back and forth motion until the two
trains collide and the fly is squashed on impact.

How far has the fly traveled, before it meets its demise?

One way to approach this problem is through infinite series. Namely, we find how far the fly during the first journey, the second journey, the third journey, etc. and add them all up. Thankfully, there is a fairly nice formula that provides this answer.

However, a much simpler way to calculate the distance is by realising that the changes in direction do not matter. Namely, all we are asking is how far can a fly travel in the hour it takes for the trains to hit each other? Clearly, this is simply 75 km. Sometimes, a moment's thought can save an hour's work!

As mentioned last time, John von Neumann was said to have immediately answered this problem, but when pressed on his solution method he said that he has used the infinite series method. Ah to have the mind of a genius!

This and other aspects of von Neumann's genius are discussed in Raymond Flood's excellent Gresham College talk, below (plus you get a bit of Alan Turing for free, which Thomas is always happy about).

Friday, 1 February 2019

A classic puzzle to start our second series. It appears in the background of A Beautiful Mind and it is said that the famous mathematician John von Neumann immediately answered with the correct result. But we'll talk about solutions later!

Two trains are on the same track. They start 100km apart and head towards each other at a speed of 50km/h.

Whilst these two trains are heading for their collision a fly starts out on the front of one train and zooms directly to the front of the other at a speed of 75km/h (see the animation above). Once the fly reaches the second train it immediately darts back to the front of the first train at the same speed and repeats this back and forth motion until the two trains collide and the fly is squashed on impact.